10-2. Parabola, Ellipse, Hyperbola
hard

જો બે ભિનન શાંકવો $x^2+y^2=4 b$ અને $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ ના છેદ બિંદુઓ, વક્ર $y^2=3 x^2$ પર આવેલા હોય, તો આ છેદ બિંદુઓ દ્વારા રચાયેલ લંબચોરસના ક્ષેત્રફળના $3 \sqrt{3}$ ઘણા........................... થાય.

A

$432$

B

$456$

C

$123$

D

$789$

(JEE MAIN-2024)

Solution

Putting $y^2=3 x^2$ in both the conics

We get $x^2=b$ and $\frac{b}{16}+\frac{3}{b}=1$

$\Rightarrow b=4,12$ (b $=4$ is rejected because curves

coincide)

$\therefore \mathrm{b}=12$

Hence points of intersection are

$( \pm \sqrt{12}, \pm 6) \Rightarrow \text { area of rectangle }=432$

Standard 11
Mathematics

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